Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void si(int &n) { scanf("%d", &n); }
void sii(int &n, int &m) { scanf("%d%d", &n, &m); }
constexpr long long int TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n - 1); }
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const int N = 1e5 + 8;
const int MOD = 1e9 + 7;
vector<pair<int, int> > ro, co, *temp;
int r, c, a[200][2000];
void go(bool earth) {
  if (!earth) {
    for (int i = 0; i < (int)(r); ++i) {
      int mn = 100000;
      for (int j = 0; j < (int)(c); ++j) mn = min(mn, a[i][j]);
      if (mn) temp->push_back({i, mn});
      for (int j = 0; j < (int)(c); ++j) a[i][j] -= mn;
    }
  } else {
    for (int i = 0; i < (int)(c); ++i) {
      int mn = 100000;
      for (int j = 0; j < (int)(r); ++j) mn = min(mn, a[j][i]);
      if (mn) temp->push_back({i, mn});
      for (int j = 0; j < (int)(r); ++j) a[j][i] -= mn;
    }
  }
}
int main() {
  sii(r, c);
  for (int i = 0; i < (int)(r); ++i)
    for (int j = 0; j < (int)(c); ++j) si(a[i][j]);
  if (r <= c) {
    temp = &ro;
    go(0);
    temp = &co;
    go(1);
  } else {
    temp = &co;
    go(1);
    temp = &ro;
    go(0);
  }
  for (int i = 0; i < (int)(r); ++i)
    for (int j = 0; j < (int)(c); ++j)
      if (a[i][j]) {
        cout << -1;
        return 0;
      }
  int ans = 0;
  for (int i = 0; i < (int)(ro.size()); ++i) ans += ro[i].second;
  for (int i = 0; i < (int)(co.size()); ++i) ans += co[i].second;
  cout << ans << endl;
  for (int i = 0; i < (int)(ro.size()); ++i)
    for (int j = 0; j < (int)(ro[i].second); ++j)
      cout << "row " << ro[i].first + 1 << endl;
  for (int i = 0; i < (int)(co.size()); ++i)
    for (int j = 0; j < (int)(co[i].second); ++j)
      cout << "col " << co[i].first + 1 << endl;
}